How do I evaluate the indefinite integral #intsin^3(x)*cos^2(x)dx# ?

1 Answer
Sep 21, 2014

The answer is #-(cos^3x)/3+(cos^5x)/5+C#.

The trick with sinusoidal powers is to use identities so that you can have #sin x# or #cos x# with a power of 1 and use substitution.

In this case, it is easier to get #sin x# to a power of 1 using #sin^2x=1-cos^2x#.

#int sin^3x*cos^2x dx#
#=int sin x(1-cos^2x)cos^2x dx#
#=int sin x(cos^2x-cos^4x)dx#
#=int sin x cos^2xdx-int sin x cos^4x dx#

Now it is a matter of using substitution:

#u=cos x#
#du = -sin x dx#

#int sin x cos^2xdx-int sin x cos^4x dx#
#=int -u^2 du+ int u^4 du#
#=-(u^3)/3+(u^5)/5+C#
#=-(cos^3x)/3+(cos^5x)/5+C#